solve for "a"...
or at least find x, y, h and k in terms of a
so
a = ""
50 e-pigs for the person who does it correctly
must use logs!!! too much work, sorry slushie
i used to be able to do these, but i havent done maths in two years, and i hated it
Also, this is probably slushbas homework but he doesn't want to do it himself :P
y=a(x-h)(x-h)+k
a(x-h)(x-h)=y+k/(x-h)(x-h)
a=(y+k)/[(x-h)(x-h)]
my quick 10 second guess
Slushba132 Wrote:solve for "a"...
or at least find x, y, h and k in terms of a
so
a = ""
50 e-pigs for the person who does it correctly
y=a(x-h)²+k
(y-k)/(x-h)²= a
too easy
by the way, that is a quadratic function
Kaiser Wrote:Slushba132 Wrote:solve for "a"...
or at least find x, y, h and k in terms of a
so
a = ""
50 e-pigs for the person who does it correctly
y=a(x-h)²+k
(y-k)/(x-h)²= a
too easy
by the way, that is a quadratic function
I WAS RIGHT
ChIcKeN BallZ Wrote:a=(y+k)/[(x-h)(x-h)]
my quick 10 second guess
(x-h)² is not the same as X²-h²
x²-h² = (x+h)(x-h)
(x-h)² = X² - 2xh + h²
but (x-h)(x-h) is the factored form to solve, First.Outer.Inner.Last.
(x*x)+(x*h)+(h*x)+(h*h)
x^2+2hx+h^2
ChIcKeN BallZ Wrote:but (x-h)(x-h) is the factored form to solve, First.Outer.Inner.Last.
you cannot factor (x-h)² except with special products, since it isn't a multiplication inside, you can't just square both terms